Finding the length of an integer in C

Question

I would like to know how I can find the length of an integer in C   For instance    1    1  25    2  12512    5 0    1   and so on   How can I do this in C

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length of n   length      i  0     1    int log10 n  1

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int digits 1   while  x gt  10       x  10      digits      return digits

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The most efficient way could possibly be to use a fast logarithm based approach  similar to those used to determine the highest bit set in an integer   size t printed length   int32 t x         size t count   x  lt  0   2   1       if   x  lt  0   x   -x       if   x  gt   100000000             count    8          x    100000000             if   x  gt   10000             count    4          x    10000             if   x  gt   100             count    2          x    100             if   x  gt   10             count       return count      This  possibly premature  optimisation takes 0 65s for 20 million calls on my netbook  iterative division like zed 0xff has takes 1 6s  recursive division like Kangkan takes 1 8s  and using floating point functions  Jordan Lewis  code  takes a whopping 6 6s  Using snprintf takes 11 5s  but will give you the size that snprintf requires for any format  not just integers  Jordan reports that the ordering of the timings are not maintained on his processor  which does floating point faster than mine   The easiest is probably to ask snprintf for the printed length    include  lt stdio h gt   size t printed length   int x         return snprintf   NULL  0    d   x       int main          int x       1  25  12512  0  -15         for   int i   0  i  lt  sizeof   x     sizeof   x 0       i           printf     d - gt   d n   x i   printed length   x i            return 0

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I think I got the most efficient way to find the length of an integer its a very simple and elegant way here it is   int PEMath  LengthOfNum int Num    int count   1     count starts at one because its the minumum amount of digits posible if  Num  lt  0        Num     -1      for int i   10  i  lt   Num  i  10         count            return count                     this loop will loop until the number  i  is bigger then  Num                     if  i  is less then  Num  multiply  i  by 10 and increase count                    when the loop ends the number of count is the length of  Num

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If you re interested in a fast and very simple solution  the following might be quickest  this depends on the probability distribution of the numbers in question    int lenHelper unsigned x        if  x  gt   1000000000  return 10      if  x  gt   100000000   return 9      if  x  gt   10000000    return 8      if  x  gt   1000000     return 7      if  x  gt   100000      return 6      if  x  gt   10000       return 5      if  x  gt   1000        return 4      if  x  gt   100         return 3      if  x  gt   10          return 2      return 1     int printLen int x        return x  lt  0   lenHelper -x    1   lenHelper x       While it might not win prizes for the most ingenious solution  it s trivial to understand and also trivial to execute - so it s fast   On a Q6600 using MSC I benchmarked this with the following loop   int res   0  for int i   -2000000000  i  lt  2000000000  i    200  res    printLen i     This solution takes 0 062s  the second-fastest solution by Pete Kirkham using a smart-logarithm approach takes 0 115s - almost twice as long   However  for numbers around 10000 and below  the smart-log is faster   At the expense of some clarity  you can more reliably beat smart-log  at least  on a Q6600    int lenHelper unsigned x            this is either a fun exercise in optimization         or it s extremely premature optimization      if x  gt   100000            if x  gt   10000000                if x  gt   1000000000  return 10              if x  gt   100000000  return 9              return 8                    if x  gt   1000000  return 7          return 6        else           if x  gt   1000                if x  gt   10000  return 5              return 4            else               if x  gt   100  return 3              if x  gt   10  return 2              return 1                      This solution is still 0 062s on large numbers  and degrades to around 0 09s for smaller numbers - faster in both cases than the smart-log approach   gcc makes faster code  0 052 for this solution and 0 09s for the smart-log approach

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Quite simple  int main         int num   123      char buf 50           convert 123 to string  buf      itoa num  buf  10           print our string     printf   s n   strlen  buf         return 0

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This goes for both negative and positive intigers      int get len int n                if n    0          return 1           if n  lt  0                           n   n    -1      if negative                    return  log10 n    1          Same logic goes for loop    int get len int n             if n    0         return 1          int len   0         if n  lt  0         n   n    -1           while n  gt  1                     n    10            len                     return len

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int get int len  int value     int l 1    while value gt 9   l    value  10      return l      and second one will work for negative numbers too   int get int len with negative too  int value     int l  value    while value   l    value  10      return l

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Why don t you cast your integer to String and get length like this   int data   123  int data len   String data  length

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The number of digits of an integer x is equal to 1   log10 x   So you can do this    include  lt math h gt   include  lt stdio h gt   int main         int x      scanf   d    amp x       printf  x has  d digits n   1    int log10 x        Or you can run a loop to count the digits yourself  do integer division by 10 until the number is 0   int numDigits   0  do         numDigits      x   x   10    while   x      You have to be a bit careful to return 1 if the integer is 0 in the first solution and you might also want to treat negative integers  work with -x if x  lt  0

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A more verbose way would be to use this function   int length int n        bool stop      int nDigits   0      int dividend   1      do               stop   false          if  n  gt  dividend                        nDigits   nDigits   1              dividend   dividend   10                    else               stop   true                        while  stop    false       return nDigits

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My way   Divide as long as number is no more divisible by 10   u8 NumberOfDigits u32 number        u8 i   1      while  number    10  i         return i      I don t know how fast is it in compared with other propositions

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Yes  using sprintf   int num  scanf   d   amp num   char testing 100   sprintf testing   d  num   int length   strlen testing     Alternatively  you can do this mathematically using the log10 function   int num  scanf   d   amp num   int length  if  num    0      length   1    else         length   log10 fabs num     1    if  num  lt  0  length

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C  Why not just take the base-10 log of the absolute value of the number  round it down  and add one  This works for positive and negative numbers that aren t 0  and avoids having to use any string conversion functions  The log10  abs  and floor functions are provided by math h  For example  int nDigits   floor log10 abs the integer      1   You should wrap this in a clause ensuring that the integer    0  since log10 0  returns -HUGE VAL according to man 3 log  Additionally  you may want to add one to the final result if the input is negative  if you re interested in the length of the number including its negative sign  Java  int nDigits   Math floor Math log10 Math abs the integer      1    N B  The floating-point nature of the calculations involved in this method may cause it to be slower than a more direct approach  See the comments for Kangkan s answer for some discussion of efficiency

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Kindly find my answer it is in one line code         include   lt stdio h gt      int main void       int c   12388884      printf  length of integer is   d  printf   d  c        return 0          that is simple and smart   Upvote if you like this

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int returnIntLength int value       int counter   0      if value  lt  0                counter            value   -value            else if value    0          return 1       while value  gt  0           value    10          counter               return counter      I think this method is well suited for this task   value and answers    -50 -  3        it will count - as one character as well if you dont want to count                 minus then remove counter   from 5th line  566666 -  6 0 -  1 505 -  3

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int main void       unsigned int n  size 0   printf  get the int     scanf   u   amp n      the magic   for int i   1  n  gt   i  i  10       size       printf  the value is   u  n   n   printf  the size is   u  n   size    return 0

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You can write a function like this   unsigned numDigits const unsigned n        if  n  lt  10  return 1      return 1   numDigits n   10

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A correct snprintf implementation   int count   snprintf NULL  0    i   x

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int intlen int integer       int a      for a   1  integer    10  a         return a

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In my opinion the shortest and easiest solution would be   int length   n   printf  Enter a number       scanf   d    amp n    length   0   while  n  gt  0       n   n   10     length       printf  Length of the number   d   length

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sprintf s    d   n   length of int   strlen s

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You may use this -   data type log10 variable name  1  ex       len    int log10 number  1

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keep dividing by ten until you get zero  then just output the number of divisions   int intLen int x      if  x  return 1    int i    for i 0  x  0    i          x    10        return i

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In this problem   i ve used some arithmetic solution   Thanks    int main void        int n  x   10  i   1      scanf  quot  d quot    amp n       while n   x  gt  0                x  10          i              printf  quot the number contains  d digits n quot   i        return 0

[c] Finding the length of an integer in C

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